Problem: N different natural numbers are given. If you select four different ones and set them as $ A $, $ B $, $ C $, $ D $, the following formula

$ \ Frac {A + B} {C --D} $



I want to find the maximum value of.




Given N different natural numbers, choose 4 different from them and create a program to find the maximum value of the above formula.



Input

The input is given in the following format.


N
a1 a2 ... aN


The number N (4 ≤ N ≤ 1000) of natural numbers is given in the first line. The value ai (1 ≤ ai ≤ 108) of each natural number is given in the second line. However, the same natural number does not appear more than once (ai ≠ aj for i ≠ j).

Output

Outputs the maximum value of the above formula as a real number for a given N natural numbers. However, the error must not exceed plus or minus 10-5.

Examples

Input

10
1 2 3 4 5 6 7 8 9 10


Output

19.00000


Input

5
22 100 42 3 86


Output

9.78947


Input

6
15 21 36 10 34 5


Output

18.00000


Input

4
100000 99999 8 1


Output

28571.285714

Solution (in CPP):

Solution:
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <complex>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <numeric>
#include <utility>
#include <iostream>
#include <algorithm>
#include <functional>

using namespace std;

#define INF (1 << 30)
#define INFL (1LL << 62)
#define MOD7 1000000007
#define MOD9 1000000009
#define EPS 1e-10

#define ll long long
#define ull unsigned long long
#define pii pair<int, int>
#define pll pair<ll, ll>
#define pul pair<ull, ull>
#define all(a) (a).begin(), (a).end()
#define mp make_pair
#define pb push_back
#define FOR(i,a,b) for (int i=(a);i<(b);++i)
#define RFOR(i,a,b) for (int i=(b)-1;i>=(a);--i)
#define REP(i,n) for (int i=0;i<(n);++i)
#define RREP(i,n) for (int i=(n)-1;i>=0;--i)

void YES() { cout << "YES" << endl; }
void NO() { cout << "NO" << endl; }
void Yes() { cout << "Yes" << endl; }
void No() { cout << "No" << endl; }
void yes() { cout << "yes" << endl; }
void no() { cout << "no" << endl; }
void Taka() { cout << "Takahashi" << endl; }
void Aoki() { cout << "Aoki" << endl; }
void digit(int d) { cout << fixed << setprecision(d); }
void IOup() { cin.tie(0); ios::sync_with_stdio(false); }

int n, a[1000];

double f1() {
	int ab = a[n - 1] + a[n - 2], cd = INF;
	for (int i = 1; i < n - 2; ++i)
		cd = min(cd, a[i] - a[i - 1]);
	return (double)ab / cd;
}

double f2() {
	int ab = 0, cd = INF, s;
	for (int i = 1; i < n; ++i) {
		if (cd > a[i] - a[i - 1]) {
			cd = a[i] - a[i - 1];
			s = i;
		}
	}
	for (int i = n - 1; i >= 0; --i) {
		if (i == s || i == s + 1 || i == s - 1)
			continue;
		ab = max(ab, a[i] + a[i - 1]);
	}
	return (double)ab / cd;
}

int main() {
	cin >> n;
	for (int i = 0; i < n; ++i)
		cin >> a[i];
	sort(a, a + n);

	cout << fixed << setprecision(6) << max(f1(), f2()) << endl;

	return 0;
}